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2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)
c: 2x-3căn x-5=0
=>2x-5căn x+2căn x-5=0
=>2căn x-5=0
=>x=25/4
Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)
a: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-\left(2\sqrt{x}+1\right)+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
b: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{x-4+5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)
2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)
\(=2\cdot\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(a,B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\left(x\ge0;x\ne1\right)\\ B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
b: Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)
\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
